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Scorpion4ik [409]
2 years ago
5

Jimmy does yard work around his neighborhood. In a given week he works at least two days, but no more than five. And on the days

he does work, he earns at least three dollars per day, but no more than seventeen dollars per day. Write an inequality that models how much money (M) Jimmy makes per week. A. 3≤M≤173≤M≤17 B. 2≤M≤52≤M≤5 C. 5≤M≤225≤M≤22 D. 6≤M≤856≤M≤85
Mathematics
1 answer:
Allushta [10]2 years ago
5 0
<span>the answer is D. 6≤M≤856≤M≤85</span>
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You sell instruments at a Caribbean music festival. You earn $326 by selling 12 sets of maracas, 6 sets of claves, and x x djemb
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326 = 12m + 6c + x = 18 rounded to 19
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There is a harvest festival each year on the planet Bozone. During that time, most of the Bozone residents sit down to rejoice a
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Answer:

The price would be decreased by 18 bozats

Step-by-step explanation:

The following information is given in the question

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Now if an extra kilogram is sold so it should be x+1

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= 300 - 18x - 18

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2 years ago
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Jill is standing at the base of her apartment building. She measures the angle of elevation to the top of a nearby tower to be 4
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1) From the measure of 40°, you can write:

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2) From the measure of 30°, you can write

tan(30°) = y / 119,18, where y is the height from the roof of Jill's building to the top of the tower.

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"Majesty Video Production Inc. wants the mean length of its advertisements to be 30 seconds. Assume the distribution of ad lengt
Westkost [7]

Answer:

a) \bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

b) Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

c) P(\bar X >31.25)=0.006=0.6\%

d) P(\bar X >28.25)=0.9997=99.97\%

e) P(28.25

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =30,\sigma =2)

We take a sample of n=16 . That represent the sample size.

a. What can we say about the shape of the distribution of the sample mean time?

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

b. What is the standard error of the mean time?

The standard error is given by this formula:

Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

c. What percent of the sample means will be greater than 31.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X >31.25)=1-P(\bar X

d. What percent of the sample means will be greater than 28.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X >28.25)=1-P(\bar X

e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?"

We want this probability:

P(28.25

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